The Molecular Basis of Inheritance

Chapter 16A. P. Biology

Mr. Knowles

Liberty Senior High

• Concept 16.1: DNA is the genetic material

• Early in the 20th century– The identification of the molecules of inheritance

loomed as a major challenge to biologists

Hammerling’s Acetabularia Exp.

• Danish biologist, Joachim Hammerling in the 1930’s.

• Used a unicellular green algae, Acetabularia.

• Proved that the hereditary material is in the nucleus.

DNA is a “Transforming Principle”

• Frederick Griffith, 1928, showed that dead bacteria could be transformed into living cells.

• Used 2 strains of Pneumococcus bacteria, one pathogenic and the other nonpathogenic.

Fig. 16.2 - Transformation

Identified the “Transforming Agent”

• Oswald Avery- in 1944, he separated and purified the different organic compounds of the bacteria and identified the DNA as responsible for the transformation effect.

Fig. 16.3

Support for Avery• Alfred Hershey and Martha Chase- in

1952, used bacteriophage (T2) as a model.

• They radiolabeled the protein coat with 35 S and the DNA with 32P in order to “see” which of the molecules actually entered the cell and produced more phage.

A Bacteriophage

Fig. 16.4: Hershey-Chase Exp.

Other Contributors• Friedrich Miescher- 1869, extracted

“nuclein”- nucleic acid from human cells and fish sperm.

• Erwin Chargaff- A = T and the G = C, called Chargaff’s Rule; equal proportion of purines and pyrimidines; amt. varied from species to species.

The Chemistry of DNA• A nucleotide = 1. PO4,

2. a five carbon sugar,

3. a nitrogen-containing base.

• Phosphodiester Bonds- 2 P-O-C bonds link nucleotides.

What is a Nucleotide? • Subunits of DNA/RNA are

Nucleotides = nitrogenous base + deoxy- or ribose sugar (5 carbons) + PO4

• Purines: Adenine and Guanine

• Pyrimidines: Cytosine, Thymine, Uracil (in RNA)

Fig. 16.8

CUT = PyAG = Pur.

Monosaccharides of Nucleic Acids

Adenosine Monophosphate

• Base = adenine• In DNA, sugar = deoxyribose

(In RNA, sugar = ribose)

• A phosphate group, PO4

• The Nucleotide = AMP

Adenosine Monophosphate

Adenosine Triphosphate (ATP)

Fig. 16.5

• Maurice Wilkins and Rosalind Franklin– Were using a technique called X-ray crystallography

to study molecular structure

• Rosalind Franklin– Produced a picture of the DNA molecule using this

technique

(a) Rosalind Franklin Franklin’s X-ray diffractionPhotograph of DNA

(b)Figure 16.6 a, b

• Franklin had concluded that DNA– Was composed of two antiparallel sugar-phosphate

backbones, with the nitrogenous bases paired in the molecule’s interior

• The nitrogenous bases– Are paired in specific combinations: adenine with

thymine, and cytosine with guanine

Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings

DNA is Antiparallel

O

–O O

OH

O

–OO

O

H2C

O

–OO

O

H2C

O

–OO

O

OH

O

O

OT A

C

GC

A T

O

O

O

CH2

OO–

OO

CH2

CH2

CH2

5 end

Hydrogen bond3 end

3 end

G

P

P

P

P

O

OH

O–

OO

O

P

P

O–

OO

O

P

O–

OO

O

P

(b) Partial chemical structure

H2C

5 endFigure 16.7b

O

Three Dimensional Structure of DNA

• Rosalind Franklin- X-ray crystallography of DNA- showed that DNA was in a helix with PO4 and sugars to the outside.

• James Watson and Francis Crick- took Franklin’s data- in April 23, 1953, and deduced the structure of DNA. Won the Nobel Prize along with Maurice Wilkins.

Fig. 16.6

Fig. 16.8

CUT = PyAG = Pur.

• Watson and Crick reasoned that there must be additional specificity of pairing

– Dictated by the structure of the bases

• Each base pair forms a different number of hydrogen bonds

– Adenine and thymine form two bonds, cytosine and guanine form three bonds

Watson and Crick

Fig. 16.7

Characteristics of DNA• All chains of DNA and RNA

have a 5’ PO4 end and a 3’ OH end.

• Base sequences are written in a 5’ to 3’ direction.

• Ex. 5’ pGpTpCpCpApT-OH 3’

Characteristics of DNA

• Base pairs stabilize the molecule by forming H-bonds.

• Antiparallel Strands-

5’----------------3’

3’----------------5’• Strands are complementary.

• In DNA replication– The parent molecule unwinds, and two new daughter

strands are built based on base-pairing rules

(a) The parent molecule has two complementary strands of DNA. Each base is paired by hydrogen bonding with its specific partner, A with T and G with C.

(b) The first step in replication is separation of the two DNA strands.

(c) Each parental strand now serves as a template that determines the order of nucleotides along a new, complementary strand.

(d) The nucleotides are connected to form the sugar-phosphate backbones of the new strands. Each “daughter” DNA molecule consists of one parental strand and one new strand.

A

C

T

A

G

A

C

T

A

G

A

C

T

A

G

A

C

T

A

G

T

G

A

T

C

T

G

A

T

C

A

C

T

A

G

A

C

T

A

G

T

G

A

T

C

T

G

A

T

C

T

G

A

T

C

T

G

A

T

C

Figure 16.9 a–d

Figure 16.10 a–c

Conservativemodel. The twoparental strandsreassociate after acting astemplates fornew strands,thus restoringthe parentaldouble helix.

Semiconservativemodel. The two strands of the parental moleculeseparate, and each functionsas a templatefor synthesis ofa new, comple-mentary strand.

Dispersivemodel. Eachstrand of bothdaughter mol-ecules containsa mixture ofold and newlysynthesizedDNA.

Parent cellFirstreplication

Secondreplication

• DNA replication is semiconservative– Each of the two new daughter molecules will have

one old strand, derived from the parent molecule, and one newly made strand

(a)

(b)

(c)

• Experiments performed by Meselson and Stahl– Supported the semiconservative model of DNA

replication

Figure 16.11

Matthew Meselson and Franklin Stahl cultured E. coli bacteria for several generations on a medium containing nucleotide precursors labeled with a heavy isotope of nitrogen, 15N. The bacteria incorporated the heavy nitrogen into their DNA. The scientists then transferred the bacteria to a medium with only 14N, the lighter, more common isotope of nitrogen. Any new DNA that the bacteria synthesized would be lighter than the parental DNA made in the 15N medium. Meselson and Stahl could distinguish DNA of different densities by centrifuging DNA extracted from the bacteria.

EXPERIMENT

The bands in these two centrifuge tubes represent the results of centrifuging two DNA samples from the flask in step 2, one sample taken after 20 minutes and one after 40 minutes.

RESULTS

Bacteriacultured inmediumcontaining15N

Bacteriatransferred tomediumcontaining14N

21

DNA samplecentrifugedafter 20 min(after firstreplication)

3 DNA samplecentrifugedafter 40 min(after secondreplication)

4Lessdense

Moredense

CONCLUSION Meselson and Stahl concluded that DNA replication follows the semiconservative model by comparing their result to the results predicted by each of the three models in Figure 16.10. The first replication in the 14N medium produced a band of hybrid (15N–14N) DNA. This result eliminated the conservative model. A second replication produced both light and hybrid DNA, a result that eliminated the dispersive model and supported the semiconservative model.

First replication Second replication

Conservativemodel

Semiconservativemodel

Dispersivemodel

Fig. 16.11- Meselson and Stahl DNA Replication Experiment

Getting Started: Origins of Replication

• The replication of a DNA molecule– Begins at special sites called origins of replication

(special sequence, AT rich), where the two strands are separated.

Bidirectional Replication in Bacteria- One Origin

Eukaryotic DNA Replication• Replicates the DNA on a chromosome in a

discrete section- Replication Unit. (about 100 kbp long).

• Prokaryotic Replication: 500 nucleotides/ second.

• Eukaryotic Replication: 50 nucleotides/second.

Problem?• If eukaryotic replication is 100 N/ sec. and

there are 3.0 X 108 N/ chromosome, how long would it take to replicate one human genome?

• Ans: 3 X 106 sec. = 34.7 days! How long does it actually take to go through S phase?

• S phase = 8 hours

• How?

Multiple Origins of Replication

• Each replication unit on a chromosome has its own origin of replication.

• Multiple units can be replicating at any given time.

• Each chromosome has numerous replication forks.

• A eukaryotic chromosome– May have hundreds or even thousands of replication

origins

Replication begins at specific siteswhere the two parental strandsseparate and form replicationbubbles.

The bubbles expand laterally, asDNA replication proceeds in bothdirections.

Eventually, the replicationbubbles fuse, and synthesis ofthe daughter strands iscomplete.

1

2

3

Origin of replication

Bubble

Parental (template) strand

Daughter (new) strand

Replication fork

Two daughter DNA molecules

In eukaryotes, DNA replication begins at many sites along the giantDNA molecule of each chromosome.

In this micrograph, three replicationbubbles are visible along the DNA ofa cultured Chinese hamster cell (TEM).

(b)(a)

0.25 µm

Figure 16.12 a, b

A eukaryotic chromosomes - have hundreds or even thousands of replication origins.

Fig. 16.12

• Elongation of new DNA at a replication fork– Is catalyzed by enzymes called DNA polymerases,

which add nucleotides to the 3 end of a growing strand

Fig. 16.13

The Problem with DNA Replication

• DNA Polymerase can only build in the 5’ to 3’ direction.

• Since the parent strands are antiparallel, the new strands are synthesized in opposite directions.

• DNA polymerases add nucleotides– Only to the free 3end of a growing strand.

• Along one template strand of DNA, the leading strand:– DNA polymerase III can synthesize a

complementary strand continuously, moving toward the replication fork

• To elongate the other new strand of DNA, the lagging strand:– DNA polymerase III must work in the direction

away from the replication fork

• The lagging strand– Is synthesized as a series of segments called

Okazaki fragments, which are then joined together by DNA ligase

Fig. 16.14

DNA Replication• Leading Stand- elongates

toward the fork, 5’ to 3’

• Lagging Strand- elongates way from fork; synthesized discontinuously in short segments-Okazaki Fragments.

Priming DNA Synthesis• DNA polymerases cannot initiate the synthesis

of a polynucleotide– They can only add nucleotides to the 3 end

• The initial nucleotide strand– Is an RNA or DNA primer made by – Primase.

• Only one primer is needed for synthesis of the leading strand– But for synthesis of the lagging strand, each

Okazaki fragment must be primed separately.

Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings

Overall direction of replication

3

3

3

35

35

35

35

35

35

35

3 5

5

1

1

21

12

5

5

12

35

Templatestrand

RNA primer

Okazakifragment

Figure 16.15

Primase joins RNA nucleotides into a primer.

1

DNA pol III adds DNA nucleotides to the primer, forming an Okazaki fragment.

2

After reaching the next RNA primer (not shown), DNA pol III falls off.

3

After the second fragment is primed. DNA pol III adds DNAnucleotides until it reaches the first primer and falls off.

4

DNA pol 1 replaces the RNA with DNA, adding to the 3 end of fragment 2.

5

DNA ligase forms a bond between the newest DNAand the adjacent DNA of fragment 1.

6 The lagging strand in this region is nowcomplete.

7

Fig. 16.16

Helicase

Single-stranded Binding (SSB) Proteins

Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings

Figure 16.16

Overall direction of replication Leadingstrand

Laggingstrand

Laggingstrand

LeadingstrandOVERVIEW

Leadingstrand

Replication fork

DNA pol III

Primase

PrimerDNA pol III Lagging

strand

DNA pol I

Parental DNA

53

43

2

Origin of replication

DNA ligase

1

5

3

Helicase unwinds theparental double helix.1

Molecules of single-strand binding proteinstabilize the unwoundtemplate strands.

2 The leading strand issynthesized continuously in the5 3 direction by DNA pol III.

3

Primase begins synthesisof RNA primer for fifthOkazaki fragment.

4

DNA pol III is completing synthesis ofthe fourth fragment, when it reaches theRNA primer on the third fragment, it willdissociate, move to the replication fork,and add DNA nucleotides to the 3 endof the fifth fragment primer.

5 DNA pol I removes the primer from the 5 endof the second fragment, replacing it with DNAnucleotides that it adds one by one to the 3 endof the third fragment. The replacement of thelast RNA nucleotide with DNA leaves the sugar-phosphate backbone with a free 3 end.

6 DNA ligase bondsthe 3 end of thesecond fragment tothe 5 end of the firstfragment.

7

• A summary of DNA replication

Replication Fork

Some Animations of DNA Replication

•http://www.wiley.com/college/pratt/0471393878/student/animations/dna_replication/index.html

- tutorial and self quiz•http://www.wiley.com/legacy/college/boyer/0470003790/animations/replication/replication.htm

- tutorial and self quiz•http://www.youtube.com/watch?v=5VefaI0LrgE&feature=related

DNA ReplicationSeveral Enzymes (+12) involved:

Helicase + ATP - unwinds the DNA and stabilizes it.

Single-stranded binding Proteins-stabilize the DNA and prevent base pairing.

Primase -makes an RNA primer.

DNA Polymerase III- builds new complementrary strands in a 5’--->3’ direction.

DNA Polymerase I – removes the primer from the 5’ end of Okazaki fragment and replaces it with DNA.

DNA Ligase – bonds the 3’ end of the second Okazaki fragment with the 5’ end of the first, and so on.

Proofreading and Repairing DNA

• DNA polymerases proofread newly made DNA– While the initial pairing errors are high as nucleotides come in.– Proofreading replaces any incorrect nucleotides; error rate of about 1

in 10 billion nucleotides.

• In mismatch repair of DNA– Repair enzymes correct errors in base pairing– Defective repair enzymes – xeroderma pigmentosum

Nucleotide Excision Repair

Fig. 16.17

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Replicating the Ends of DNA Molecules

• The ends of eukaryotic chromosomal DNA

– Get shorter with each round of replication

Figure 16.18

End of parentalDNA strands

Leading strandLagging strand

Last fragment Previous fragment

RNA primer

Lagging strand

Removal of primers andreplacement with DNAwhere a 3 end is available

Primer removed butcannot be replacedwith DNA becauseno 3 end available

for DNA polymerase

Second roundof replication

New leading strand

New lagging strand 5

Further roundsof replication

Shorter and shorterdaughter molecules

5

3

5

3

5

3

5

3

3

Fig. 16.18

• Eukaryotic chromosomal DNA molecules– Have at their ends nucleotide sequences that repeat

(100 – 1,000), called telomeres, that postpone the erosion of genes near the ends of DNA molecules

Figure 16.19 1 µm

Show me the Replication!

DNA Packaging

• Eukaryotic DNA is packaged into nucleosomes- 146 bp of DNA wrapped around histones.

• Condenses the DNA into a nucleus.

• Protects the DNA from exposure.

How many genes are on a chromosome?

• Human Chromosome #22- composed of 33.4 megabases (Mb), identified 545 genes, 298 of these were unknown.

• 100-200 more genes may be identified!

• The rest is “junk” DNA.

Chromosomes have...

• Humans have about 3 X 109 bp of DNA.

• Conservative estimate is 30,000 genes.

• Therefore, the average gene is 100,000 bp long.

One Gene-One Enzyme Hypothesis

• 1941- Beadle and Tatum- created mutants in the fungus Neurospora using X-rays.

• Exposed wild-type spores to the mutagen.

• Change growth from a complete to minimal media.

• Mutants no longer able to synthesize essential organics (e.g. amino acids)

One Gene-One Enzyme• Beadle and Tatum- found several mutants.

• Found a different site for each enzyme.

• Each mutant had a different mutation at a different site (enzyme) on the chromosome.

• Each mutant had a defect in a different enzyme.

• Concluded: one gene encodes one enzyme.

One Gene-One Enzyme Hypothesis

• Genetic traits are expressed as a result of the activities of enzymes.

• Many enzymes contain multiple subunits.

• Each subunit encoded by a differernt gene.

• Now, referred to as one gene-one polypeptide.

How does DNA Encode Proteins?

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Some Differences between Prokaryotic and

Eukaryotic Protein Synthesis

Prokaryotes Eukaryotes• Lack introns- no

mRNA processing.

• Begin translation before transcription is finished.

• Have introns- mRNA is spliced.

• mRNA completely formed before translation.

• mRNA has a 5’ methyl G cap added

Prokaryotes Eukaryotes• mRNA begins at

an AUG start codon, no cap.

• Multiple genes on one mRNA.

• 70S ribosome used.

• mRNA has a poly A tail added at the 3’.

• A single gene on one mRNA.

• 80S ribosome used.

Other Differences in Gene Organization• Tandem Clusters- several hundred genes

organized together; Ex. rRNA genes.

• Multigene Families- genes very different from each other, but encode similar proteins; Ex. globin genes.

• Pseudogenes- silent copies of genes inactivated by mutations.

Transposons

Acute Promyelocytic Leukemia- Several

Translocation Events; one is between #7 and #15.

DNA ReplicationProkaryotes

• single, circular chromosome.

• one single origin on the chromosome.

• rate of over 1,000 nt/second.

Eukaryotes

• multiple, linear chromosomes.

• several origins on each chromosome.

• rate of about 50-100 nt/second